While solving a problem related to electric fields I came across a conclusion in the solution that, for a "straight, steady current carrying conducting wire" having current $I$, so that the charges flow on its surface (obviously because it's a conductor), the electric field just outside the wire $E_{out} \neq 0$ i.e. finite and electric field just inside $E_{in}= 0$. (Note $E_{in}$ does not mean $E$ field at any point inside wire - it's for a point just inside the surface of the wire, same for $ E_{out}.$)
But I think it should be just opposite because just outside the wire or be it far away from wire, when the wire is electrically neutral, it would have $E_{out}=0$. I know that $\vec E$ just outside a conductor is given as $\frac{\sigma}{2\epsilon_0}$ but that is for the case when the conductor has a net charge. But the wire here should be neutral.
For the second part, considering the wire to be cylindrical, the negative charge that comes to the surface forms a shell, and $E$ field due to the symmetric shell charge is 0. But then the positive immobile ions in the body of wire would exert a net positive $E$ field just inside the wire so $E_{in} \neq 0$.
The Original Problem-
Q.19 A steady current in a straight conducting wire produces a surface charge on it. Let $E_{out}$ and $E_{in }$be the magnitudes of the electric fields just outside and just inside the wire, respectively. Which of the following statements is true for these fields?
(a) Eout is always greater than Ein
(b) Eout is always smaller than Ein
(c) Eout could be greater or smaller than Ein
(d) Eout is equal to Ein
Solution: (a)
In this case, $E_{in}=0$, $E_{out} \neq 0$. So $E_{out}$ > $E_{in}$
